# Solution Review: Tail Recursion

In the following lesson, we will go over the solution of the challenge: Tail Recursion.

We'll cover the following

In this challenge, you had to create a tail-recursive factorial function.

## Solution

A skeleton of the function was already provided for you. Let’s look it over.

def factorial(x: Int): Int = {
def loop(accumulator: Int, x: Int): Int = {

}
loop(1,x)
}


As before, factorial takes a single parameter x of type Int. However, this time, its function body consists of a nested function loop. loop is the tail recursive part of factorial. It has two parameters accumulator and x. The accumulator stores the current value of the factorial in each recursive call. This is why when we pass 1 as the initial accumulator when loop is called in the function body of factorial.

loop(1,x)


1 is the smallest possible factorial.

Hence, if the number whose factorial we want to find is 0, we will simply return the accumulator as is, this is our base case.

if(x == 0) accumulator


The recursive case is if x is not equal to 0. We will recursively call loop in this case. The new accumulator to be passed will be accumulator * x and the new x to be passed will be x-1.

else loop(accumulator*x,x-1)


As the last thing being done by loop is a recursive call, it is a tail-recursive function.

You can find the complete solution below:

You were required to write the code on line 3 and line 4.